1. Suppose 1.00 g of NaOH is used to prepare 250 mL of an NaOH solution. Compare the expected molarity of this solution to the a
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Answer : A= 0.5, B = 0.25 , C = 0.125, D = 0.015625 and E = 0.00390625
Explanation :
Half life of a substance is defined as the amount of time taken by the substance to reduce to half of its original amount.
Here n represents the number of half lives.
The amount of substance that remains after n half lives can be calculated using the given formula,
So when we have n =1,
Fraction of substance that remains = 0.5¹ = 0.5.
That means after first half life over, the amount of substance that remains is 0.5 times that of original.
Therefore we have A = 0.5
When n = 2, we have 0.5² = 0.25
So when 2 half lives are over, the amount of substance that remains is 0.25 times that of original
Therefore B = 0.25
When n = 3, we have 0.5³ = 0.125
So when 3 half lives are over, the amount of substance that remains is 0.125 times that of original.
Therefore we have C = 0.125
When n = 6 , we have 0.5⁶ = 0.015625
So D = 0.015625
When n = 8, we have 0.5⁸ = 0.00390625
Therefore E = 0.00390625
The values for A, B, C, D and E are 0.5, 0.25, 0.125, 0.015625 and 0.00390625 respectively.